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Leibniz notation clarification?

  1. Sep 23, 2009 #1
    dy = lim [tex]\Delta[/tex]x-->0 (f(x+[tex]\Delta[/tex]x) - f(x))

    dx = lim [tex]\Delta[/tex]x-->0 ([tex]\Delta[/tex]x)

    Therefore dy/dx is f'(x) if f(x) = y

    Is all of this true? I'm tired of integrating with variable substitution and not knowing what du by itself really means. People are always saying that Leibniz notation doesn't literally represent a fraction, and the seemingly algebraic manipulation is more complicated than it looks. But can't anyone say what exactly is going on? Maybe give me the rigorous definition of "dy" if I was wrong. Thanks.

    Oh, and if the above is correct, then can we say that dy/dx is a fraction of limits?
     
  2. jcsd
  3. Sep 23, 2009 #2
    Both of those limits are zero.

    dy/dx is 0/0 in your guess, not useful by itself

    No

    Some calculus textbooks invent meanings for it. But the meaning of du by itself used in modern mathematics has to wait for differential topology, a more advanced subject.

     
  4. Sep 23, 2009 #3

    lurflurf

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    The meaning of dy depends on context. In most introductory calculus books dy would be defined by
    dy=f'(x)dx
    dy is a linearization of Δy
    with this definition it is correct to regard dy/dx as a fraction with
    dy/dx=f'(x)dx/dx=f'(x)
     
  5. Sep 24, 2009 #4
    Thanks for the help guys; I've been throwing my limit properties around too carelessly! :rolleyes: Well, live and learn...

    Differential Topology huh? I guess I'll stick with visualizing infinitesimals until then, like Leibniz, but always remembering that it's not the final word. Hopefully that won't lead me far astray.

    What's dx then? Seems circular to me. Probably everything will be revealed when I have a bare-bones definition of differentials (maybe using differential topology.)


    By the way, ever heard the Simpsons joke, where the derivative of r3/3 is r2*dr, or "r*dr*r" (hardy-har-har)? This must be wrong, because that's not the derivative, it's the differential of r3/3 or something, right? I used to think this joke was pretty funny, but now I hate it, because it confused me a little. :wink:

    Or am I confused now? :confused:
     
  6. Sep 24, 2009 #5

    lurflurf

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    It is not circular, but it does not allow one to do the wacky stuff like proving the chain rule by cancellation. To do the wacky stuff you need nonstandard analysis which requires a different foundation than standard analysis including logic and ultrafilters. You may be disappointed to know that the differential geometry definition of differential is not so different from the one given in elementaryy calculus. That is true infinitesimals are not used, but such definitions are closely connected to what the differential is, a linearization of the difference. There is also an algebraic definition using dual numbers.

    The idea of a differential defined in elementary calculus as
    dy=f'(x)dx
    is we want the linear function closest to f near x
    y=mx+b
    we see that
    m=f'(x)
    b=f(x)-xf'(x)
    we chose new coordinates with origin (x,f(x)) so the equation is
    dy=f(x)dx
    we now have a function that looks everywhere how f looks near (x,f(x))
    we also sometimes use it to approximate the difference
    dy~Δy
    f'(x)dx~f(x+dx)-f(x)
     
    Last edited: Sep 24, 2009
  7. Sep 25, 2009 #6
    Having a strong, sturdy understanding of differentials must be key to mastering calculus, so I'm not going to stop til I get there!

    Let me get this straight.

    Take a function and a line tangent to it at x, and let a change in y be "Δy" and a change in y on the tangent line be "Δylin"=f'(x)Δx

    and Δy~Δylin

    which means you can approximate a change in y linearly with f'(x)Δx

    so a differential means that dy=Δy=Δylin=f'(x)dx.
    A differential is so infinitesimally small that it makes the linear approximation an "equality".

    Is that it?

    P.S. How do you write "Δ"? (I copy and pasted that)
     
  8. Sep 26, 2009 #7

    lurflurf

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    A better way to think about it is that the differential picks oup an approximation. Once it has done the picking we need not and perhaps should not regard the differential as being small.
    example
    d(sin(x))=cos(x)dx
    is saying
    let us pick out the fuction that behaves everywhere how [sin(x+dx)-sin(x)] behaves when dx is "small"
    ok here it is cos(x)dx
    once this is done it is no longer required that dx is small
    we might have x=1 dx=1 so
    d(sin(x))|x,dx=1=cos(x)dx|x,dx=1=cos(1)~0.54030230586813971740093660744298
    even though
    Δ(sin(x))|x,dx=1=sin(x+dx)-sin(x)=sin(2)-sin(1)~0.067826442017785188743517544281446
    thus the approximation would not be considered good in many application
    or x=1 dx=10^100
    d(sin(x))|x=1,dx=10^100=cos(x)dx|x=1,dx=10^100=cos(1)*10^100
    which is even worse
     
    Last edited: Sep 26, 2009
  9. Sep 26, 2009 #8
    So basically, the differential starts out as a linear approximation, as I said earlier, only it doesn't end by turning the approximation into an equality?

    So far, so good.

    Instead, it may conclude with dy=Δy ~ Δylin=f'(x)dx

    Meaning dy is the actual change in y, which can be estimated linearly with f'(x)dx.
    Or is dy=Δylin, which then estimates the real change in y?

    And is this truly the rigorous, mathematical definition? It sounds more like the physical interpretation, where a differential is simply smaller than can be measured by the instrument. While this interpretation will likely form the basis of my intuition, right now I'm curious about the abstract formality.

    Your reference to applications and arbitrarily assigned values of dx suggested to me the physical sciences.
     
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